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Let $X$ be a smooth projective $k$-scheme, where $k=\mathbb{F}_p$ and $p$ is prime. We have an identification of the de Rham cohomology of $X$ with $H^*_{crys}(X/k)$: $H^*_{DR}(X/k)\cong H^*_{crys}(X/k)$ (note that I am not considering crystalline cohomology $H^*_{crys}(X/W)=\lim H^*_{crys}(X/W_n)$, where $W=\mathbb{Z}_p$ is the Witt ring of $k$, and $W_n=W/p^n$). My understanding is that $H^*_{crys}(X/k)$ is the cohomology of a sheaf of $k$-algebras $\mathcal{O}$ on a site. Thus, we may consider the Frobenius endomorphism of $\mathcal{O}$: $F:\mathcal{O}\to \mathcal{O}$, $t\mapsto t^p$ for every section $t$. Clearly, $F$ induces a linear (because $k=\mathbb{F}_p$) endomorphism of any injective resolution of $\mathcal{O}$, and so we get an induced endomorphism $F^*$ of $H^*_{DR}(X/k)$.

Questions: is $F^*$ always the identity? If not, how to compute $F^*$? For example, what is $F^*$ if $X$ is a projective space (other examples would be welcome too)?

I tried to find this information in standard references, but they always consider the case of $X/\mathbb{Z}_p$, and they consider $H^*_{crys}(X/W)[1/p]$.

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    $\begingroup$ On $\mathbf P^n$, the map is just $[x_0:\ldots:x_n] \mapsto [x_0^p:\ldots:x_n^p]$. In arbitrary characteristic, the map $[x_0:\ldots:x_n] \mapsto [x_0^d:\ldots:x_n^d]$ acts on $H^i(\mathbf P^n,\Omega^i)$ as multiplication by $d^i$, so in this case its zero for $i > 0$. $\endgroup$ Commented Jun 11, 2020 at 20:59
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    $\begingroup$ Another example: if $E$ is an elliptic curve, then the Frobenius action on $H^1(E,\mathcal O_E)$ is $0$ if and only if $E$ is supersingular. More generally, it is related to de Rham–Witt cohomology, which rationally gives the slope filtration on crystalline cohomology. So the answer depends on subtle arithmetic information. $\endgroup$ Commented Jun 11, 2020 at 20:59
  • $\begingroup$ Thank you! Do you have a reference which explains the relation to de Rham–Witt cohomology? $\endgroup$
    – Vitay
    Commented Jun 11, 2020 at 22:04
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    $\begingroup$ Expanding on what was already said: the Frobenius map $H^*_{dR}(X) \to H^*_{dR}(X)$ factors as $H^*_{dR}(X) \to H^*(X,O_X) \to H^*(X,O_X) \to H^*_{dR}(X)$ where the first and last map are edge maps for the Hodge/conjugate spectral sequences, and the middle map is the Frobenius on structure sheaf cohomology. So understanding, e.g., the rank of the Frobenius on de Rham cohomology requires knowing that rank of the Frobenius on $O_X$-cohomology as well as behaviour of certain differentials in the Hodge/conjugate spectral sequences. $\endgroup$
    – Anonymous
    Commented Jun 11, 2020 at 22:43

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